我模拟了随机数据集与给定的特征和一切工作只是罚款:
df <- data.frame(Subject = c("Sub1", "Sub2", "Sub3", "Sub4", "Sub5", "Sub6", "Sub7", "Sub8", "Sub9", "Sub10", "Sub11", "Sub12", "Sub13", "Sub14", "Sub15", "Sub16", "Sub17", "Sub18", "Sub19", "Sub20", "Sub21", "Sub22", "Sub23", "Sub24", "Sub25", "Sub26", "Sub27", "Sub28", "Sub29", "Sub30"),
Disability = c("0", "0", "1", "1", "1", "1", "0", "0", "0", "1", "1", "0", "0", "0", "0", "1", "0", "0", "1", "0", "0", "0", "0", "1", "1", "1", "0", "0", "1", "0"),
Pref = c("touchpad", "touchpad", "touchpad", "trackball", "trackball", "trackball", "trackball", "trackball", "trackball", "trackball", "trackball", "trackball", "touchpad", "trackball", "trackball", "touchpad", "touchpad", "trackball", "touchpad", "trackball", "touchpad", "touchpad", "trackball", "touchpad", "touchpad", "touchpad", "touchpad", "touchpad", "trackball", "trackball"))
给定的命令的结果是以下
binom.test(sum(df[df$Disability == "0",]$Pref == "touchpad"),
nrow(df[df$Disability == "0",]), p=1/2)
Exact binomial test
data: sum(df[df$Disability == "0", ]$Pref == "touchpad") and nrow(df[df$Disability == "0", ])
number of successes = 8, number of trials = 18, p-value = 0.8145
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
0.2153015 0.6924283
sample estimates:
probability of success
0.4444444
编辑
为了将相同的测试应用于真实数据(链接到评论中给出的文件),第一步应当由命令读出存储在实际数据帧中的值来替换:
df <- read.csv("deviceprefs-1.csv")
另外,给出的命令执行二项式检验工作得很好与真实数据组。
如果您提供了您正在使用的数据,那会更好,因此我们可以重现您的代码。试试'dput',或者在某处上传这个csv并发布一个链接。 –
也请在问题中添加错误消息。 –
感谢您的帮助!我只是想出了我需要用构建的xtab的名称替换“df”。文件:https://www.dropbox.com/s/rd796wor7by5uky/DesignExperiments_R.Rproj?dl=0 https://www.dropbox.com/s/cig2u4d5vpkjma1/deviceprefs.csv?dl = 0 – testimo